# m Coloring Problem | Backtracking-5

Given an undirected graph and a number m, determine if the graph can be coloured with at most m colours such that no two adjacent vertices of the graph are colored with the same color. Here coloring of a graph means the assignment of colors to all vertices.
Input-Output format:
Input:

1. A 2D array graph[V][V] where V is the number of vertices in graph and graph[V][V] is adjacency matrix representation of the graph. A value graph[i][j] is 1 if there is a direct edge from i to j, otherwise graph[i][j] is 0.
2. An integer m which is the maximum number of colors that can be used.

Output:
An array color[V] that should have numbers from 1 to m. color[i] should represent the color assigned to the ith vertex. The code should also return false if the graph cannot be colored with m colors.

Example:

```Input:
graph = {0, 1, 1, 1},
{1, 0, 1, 0},
{1, 1, 0, 1},
{1, 0, 1, 0}
Output:
Solution Exists:
Following are the assigned colors
1  2  3  2
Explanation: By coloring the vertices
vertices does not have same colors

Input:
graph = {1, 1, 1, 1},
{1, 1, 1, 1},
{1, 1, 1, 1},
{1, 1, 1, 1}
Output: Solution does not exist.
Explanation: No solution exits.
```

Following is an example of a graph that can be coloured with 3 different colours. ## We strongly recommend that you click here and practice it, before moving on to the solution.

Method 1: Naive.
Naive Approach: Generate all possible configurations of colours. Since each node can be coloured using any of the m available colours, the total number of colour configurations possible are m^V.
After generating a configuration of colour, check if the adjacent vertices have the same colour or not. If the conditions are met, print the combination and break the loop.

Algorithm:

1. Create a recursive function that takes current index, number of vertices and output color array.
2. If the current index is equal to number of vertices. Check if the output color configuration is safe, i.e check if the adjacent vertices does not have same color. If the conditions are met, print the configuration and break.
3. Assign color to a vertex (1 to m).
4. For every assigned color recursively call the function with next index and number of vertices
5. If any recursive function returns true break the loop and return true.

 `#include ` `#include ` ` `  `// Number of vertices in the graph ` `#define V 4 ` ` `  `void` `printSolution(``int` `color[]); ` ` `  `// check if the colored ` `// graph is safe or not ` `bool` `isSafe( ` `    ``bool` `graph[V][V], ``int` `color[]) ` `{ ` `    ``// check for every edge ` `    ``for` `(``int` `i = 0; i < V; i++) ` `        ``for` `(``int` `j = i + 1; j < V; j++) ` `            ``if` `( ` `                ``graph[i][j] && color[j] == color[i]) ` `                ``return` `false``; ` `    ``return` `true``; ` `} ` ` `  `/* This function solves the m Coloring  ` `   ``problem using recursion. It returns ` `  ``false if the m colours cannot be assigned,  ` `  ``otherwise, return true and prints  ` `  ``assignments of colours to all vertices.  ` `  ``Please note that there may be more than  ` `  ``one solutions, this function prints one  ` `  ``of the feasible solutions.*/` `bool` `graphColoring( ` `    ``bool` `graph[V][V], ``int` `m, ` `    ``int` `i, ``int` `color[V]) ` `{ ` `    ``// if current index reached end ` `    ``if` `(i == V) { ` `        ``// if coloring is safe ` `        ``if` `(isSafe(graph, color)) { ` `            ``// Print the solution ` `            ``printSolution(color); ` `            ``return` `true``; ` `        ``} ` `        ``return` `false``; ` `    ``} ` ` `  `    ``// Assign each color from 1 to m ` `    ``for` `(``int` `j = 1; j <= m; j++) { ` `        ``color[i] = j; ` ` `  `        ``// Recur of the rest vertices ` `        ``if` `(graphColoring( ` `                ``graph, m, i + 1, color)) ` `            ``return` `true``; ` ` `  `        ``color[i] = 0; ` `    ``} ` ` `  `    ``return` `false``; ` `} ` ` `  `/* A utility function to print solution */` `void` `printSolution(``int` `color[]) ` `{ ` `    ``printf``( ` `        ``"Solution Exists:"` `        ``" Following are the assigned colors \n"``); ` `    ``for` `(``int` `i = 0; i < V; i++) ` `        ``printf``(``" %d "``, color[i]); ` `    ``printf``(``"\n"``); ` `} ` ` `  `// Driver program to test above function ` `int` `main() ` `{ ` `    ``/* Create following graph and  ` `       ``test whether it is 3 colorable ` `      ``(3)---(2) ` `       ``|   / | ` `       ``|  /  | ` `       ``| /   | ` `      ``(0)---(1) ` `    ``*/` `    ``bool` `graph[V][V] = { ` `        ``{ 0, 1, 1, 1 }, ` `        ``{ 1, 0, 1, 0 }, ` `        ``{ 1, 1, 0, 1 }, ` `        ``{ 1, 0, 1, 0 }, ` `    ``}; ` `    ``int` `m = 3; ``// Number of colors ` ` `  `    ``// Initialize all color values as 0. ` `    ``// This initialization is needed ` `    ``// correct functioning of isSafe() ` `    ``int` `color[V]; ` `    ``for` `(``int` `i = 0; i < V; i++) ` `        ``color[i] = 0; ` ` `  `    ``if` `(!graphColoring( ` `            ``graph, m, 0, color)) ` `        ``printf``(``"Solution does not exist"``); ` ` `  `    ``return` `0; ` `} `

Output:

```Solution Exists: Following are the assigned colors
1  2  3  2
```

Complexity Analysis:

• Time Complexity: O(m^V).
There are total O(m^V) combination of colors. So the time complexity is O(m^V).
• Space Complexity: O(V).
To store the output array O(V) space is required.

Method 2: Backtracking.

Approach: The idea is to assign colors one by one to different vertices, starting from the vertex 0. Before assigning a color, check for safety by considering already assigned colors to the adjacent vertices i.e check if the adjacent vertices have the same color or not. If there is any color assignment that does not violate the conditions, mark the color assignment as part of the solution. If no assignment of color is possible then backtrack and return false.

Algorithm:

1. Create a recursive function that takes the graph, current index, number of vertices and output color array.
2. If the current index is equal to number of vertices. Print the color configuration in output array.
3. Assign color to a vertex (1 to m).
4. For every assigned color, check if the configuration is safe, (i.e. check if the adjacent vertices do not have the same color) recursively call the function with next index and number of vertices
5. If any recursive function returns true break the loop and return true.
6. If no recusive function returns true then return false.

## C/C++

 `#include ` `#include ` ` `  `// Number of vertices in the graph ` `#define V 4 ` ` `  `void` `printSolution(``int` `color[]); ` ` `  `/* A utility function to check if  ` `   ``the current color assignment ` `   ``is safe for vertex v i.e. checks  ` `   ``whether the edge exists or not ` `   ``(i.e, graph[v][i]==1). If exist  ` `   ``then checks whether the color to  ` `   ``be filled in the new vertex(c is ` `   ``sent in the parameter) is already ` `   ``used by its adjacent  ` `   ``vertices(i-->adj vertices) or  ` `   ``not (i.e, color[i]==c) */` `bool` `isSafe( ` `    ``int` `v, ``bool` `graph[V][V], ` `    ``int` `color[], ``int` `c) ` `{ ` `    ``for` `(``int` `i = 0; i < V; i++) ` `        ``if` `( ` `            ``graph[v][i] && c == color[i]) ` `            ``return` `false``; ` `    ``return` `true``; ` `} ` ` `  `/* A recursive utility function  ` `to solve m coloring problem */` `bool` `graphColoringUtil( ` `    ``bool` `graph[V][V], ``int` `m, ` `    ``int` `color[], ``int` `v) ` `{ ` `    ``/* base case: If all vertices are  ` `       ``assigned a color then return true */` `    ``if` `(v == V) ` `        ``return` `true``; ` ` `  `    ``/* Consider this vertex v and  ` `       ``try different colors */` `    ``for` `(``int` `c = 1; c <= m; c++) { ` `        ``/* Check if assignment of color  ` `           ``c to v is fine*/` `        ``if` `(isSafe( ` `                ``v, graph, color, c)) { ` `            ``color[v] = c; ` ` `  `            ``/* recur to assign colors to  ` `               ``rest of the vertices */` `            ``if` `( ` `                ``graphColoringUtil( ` `                    ``graph, m, color, v + 1) ` `                ``== ``true``) ` `                ``return` `true``; ` ` `  `            ``/* If assigning color c doesn't ` `               ``lead to a solution then remove it */` `            ``color[v] = 0; ` `        ``} ` `    ``} ` ` `  `    ``/* If no color can be assigned to  ` `       ``this vertex then return false */` `    ``return` `false``; ` `} ` ` `  `/* This function solves the m Coloring  ` `   ``problem using Backtracking. It mainly  ` `   ``uses graphColoringUtil() to solve the  ` `   ``problem. It returns false if the m  ` `   ``colors cannot be assigned, otherwise  ` `   ``return true and prints assignments of  ` `   ``colors to all vertices. Please note  ` `   ``that there may be more than one solutions, ` `   ``this function prints one of the ` `   ``feasible solutions.*/` `bool` `graphColoring( ` `    ``bool` `graph[V][V], ``int` `m) ` `{ ` `    ``// Initialize all color values as 0. ` `    ``// This initialization is needed ` `    ``// correct functioning of isSafe() ` `    ``int` `color[V]; ` `    ``for` `(``int` `i = 0; i < V; i++) ` `        ``color[i] = 0; ` ` `  `    ``// Call graphColoringUtil() for vertex 0 ` `    ``if` `( ` `        ``graphColoringUtil( ` `            ``graph, m, color, 0) ` `        ``== ``false``) { ` `        ``printf``(``"Solution does not exist"``); ` `        ``return` `false``; ` `    ``} ` ` `  `    ``// Print the solution ` `    ``printSolution(color); ` `    ``return` `true``; ` `} ` ` `  `/* A utility function to print solution */` `void` `printSolution(``int` `color[]) ` `{ ` `    ``printf``( ` `        ``"Solution Exists:"` `        ``" Following are the assigned colors \n"``); ` `    ``for` `(``int` `i = 0; i < V; i++) ` `        ``printf``(``" %d "``, color[i]); ` `    ``printf``(``"\n"``); ` `} ` ` `  `// driver program to test above function ` `int` `main() ` `{ ` `    ``/* Create following graph and test  ` `       ``whether it is 3 colorable ` `      ``(3)---(2) ` `       ``|   / | ` `       ``|  /  | ` `       ``| /   | ` `      ``(0)---(1) ` `    ``*/` `    ``bool` `graph[V][V] = { ` `        ``{ 0, 1, 1, 1 }, ` `        ``{ 1, 0, 1, 0 }, ` `        ``{ 1, 1, 0, 1 }, ` `        ``{ 1, 0, 1, 0 }, ` `    ``}; ` `    ``int` `m = 3; ``// Number of colors ` `    ``graphColoring(graph, m); ` `    ``return` `0; ` `} `

## Java

 `/* Java program for solution of  ` `   ``M Coloring problem using backtracking */` `public` `class` `mColoringProblem { ` `    ``final` `int` `V = ``4``; ` `    ``int` `color[]; ` ` `  `    ``/* A utility function to check  ` `       ``if the current color assignment  ` `       ``is safe for vertex v */` `    ``boolean` `isSafe( ` `        ``int` `v, ``int` `graph[][], ``int` `color[], ` `        ``int` `c) ` `    ``{ ` `        ``for` `(``int` `i = ``0``; i < V; i++) ` `            ``if` `( ` `                ``graph[v][i] == ``1` `&& c == color[i]) ` `                ``return` `false``; ` `        ``return` `true``; ` `    ``} ` ` `  `    ``/* A recursive utility function  ` `       ``to solve m coloring  problem */` `    ``boolean` `graphColoringUtil( ` `        ``int` `graph[][], ``int` `m, ` `        ``int` `color[], ``int` `v) ` `    ``{ ` `        ``/* base case: If all vertices are  ` `           ``assigned a color then return true */` `        ``if` `(v == V) ` `            ``return` `true``; ` ` `  `        ``/* Consider this vertex v and try  ` `           ``different colors */` `        ``for` `(``int` `c = ``1``; c <= m; c++) { ` `            ``/* Check if assignment of color c to v ` `               ``is fine*/` `            ``if` `(isSafe(v, graph, color, c)) { ` `                ``color[v] = c; ` ` `  `                ``/* recur to assign colors to rest ` `                   ``of the vertices */` `                ``if` `( ` `                    ``graphColoringUtil( ` `                        ``graph, m, ` `                        ``color, v + ``1``)) ` `                    ``return` `true``; ` ` `  `                ``/* If assigning color c doesn't lead ` `                   ``to a solution then remove it */` `                ``color[v] = ``0``; ` `            ``} ` `        ``} ` ` `  `        ``/* If no color can be assigned to  ` `           ``this vertex then return false */` `        ``return` `false``; ` `    ``} ` ` `  `    ``/* This function solves the m Coloring problem using ` `       ``Backtracking. It mainly uses graphColoringUtil() ` `       ``to solve the problem. It returns false if the m ` `       ``colors cannot be assigned, otherwise return true ` `       ``and  prints assignments of colors to all vertices. ` `       ``Please note that there  may be more than one ` `       ``solutions, this function prints one of the ` `       ``feasible solutions.*/` `    ``boolean` `graphColoring(``int` `graph[][], ``int` `m) ` `    ``{ ` `        ``// Initialize all color values as 0. This ` `        ``// initialization is needed correct ` `        ``// functioning of isSafe() ` `        ``color = ``new` `int``[V]; ` `        ``for` `(``int` `i = ``0``; i < V; i++) ` `            ``color[i] = ``0``; ` ` `  `        ``// Call graphColoringUtil() for vertex 0 ` `        ``if` `( ` `            ``!graphColoringUtil( ` `                ``graph, m, color, ``0``)) { ` `            ``System.out.println( ` `                ``"Solution does not exist"``); ` `            ``return` `false``; ` `        ``} ` ` `  `        ``// Print the solution ` `        ``printSolution(color); ` `        ``return` `true``; ` `    ``} ` ` `  `    ``/* A utility function to print solution */` `    ``void` `printSolution(``int` `color[]) ` `    ``{ ` `        ``System.out.println( ` `            ``"Solution Exists: Following"` `            ``+ ``" are the assigned colors"``); ` `        ``for` `(``int` `i = ``0``; i < V; i++) ` `            ``System.out.print(``" "` `+ color[i] + ``" "``); ` `        ``System.out.println(); ` `    ``} ` ` `  `    ``// driver program to test above function ` `    ``public` `static` `void` `main(String args[]) ` `    ``{ ` `        ``mColoringProblem Coloring  ` `= ``new` `mColoringProblem(); ` `        ``/* Create following graph and  ` `           ``test whether it is ` `           ``3 colorable ` `          ``(3)---(2) ` `           ``|   / | ` `           ``|  /  | ` `           ``| /   | ` `          ``(0)---(1) ` `        ``*/` `        ``int` `graph[][] = { ` `            ``{ ``0``, ``1``, ``1``, ``1` `}, ` `            ``{ ``1``, ``0``, ``1``, ``0` `}, ` `            ``{ ``1``, ``1``, ``0``, ``1` `}, ` `            ``{ ``1``, ``0``, ``1``, ``0` `}, ` `        ``}; ` `        ``int` `m = ``3``; ``// Number of colors ` `        ``Coloring.graphColoring(graph, m); ` `    ``} ` `} ` `// This code is contributed by Abhishek Shankhadhar `

## Python

 `# Python program for solution of M Coloring  ` `# problem using backtracking ` ` `  `class` `Graph(): ` ` `  `    ``def` `__init__(``self``, vertices): ` `        ``self``.V ``=` `vertices ` `        ``self``.graph ``=` `[[``0` `for` `column ``in` `range``(vertices)]\ ` `                              ``for` `row ``in` `range``(vertices)] ` ` `  `    ``# A utility function to check if the current color assignment ` `    ``# is safe for vertex v ` `    ``def` `isSafe(``self``, v, colour, c): ` `        ``for` `i ``in` `range``(``self``.V): ` `            ``if` `self``.graph[v][i] ``=``=` `1` `and` `colour[i] ``=``=` `c: ` `                ``return` `False` `        ``return` `True` `     `  `    ``# A recursive utility function to solve m ` `    ``# coloring  problem ` `    ``def` `graphColourUtil(``self``, m, colour, v): ` `        ``if` `v ``=``=` `self``.V: ` `            ``return` `True` ` `  `        ``for` `c ``in` `range``(``1``, m ``+` `1``): ` `            ``if` `self``.isSafe(v, colour, c) ``=``=` `True``: ` `                ``colour[v] ``=` `c ` `                ``if` `self``.graphColourUtil(m, colour, v ``+` `1``) ``=``=` `True``: ` `                    ``return` `True` `                ``colour[v] ``=` `0` ` `  `    ``def` `graphColouring(``self``, m): ` `        ``colour ``=` `[``0``] ``*` `self``.V ` `        ``if` `self``.graphColourUtil(m, colour, ``0``) ``=``=` `None``: ` `            ``return` `False` ` `  `        ``# Print the solution ` `        ``print` `"Solution exist and Following are the assigned colours:"` `        ``for` `c ``in` `colour: ` `            ``print` `c, ` `        ``return` `True` ` `  `# Driver Code ` `g ``=` `Graph(``4``) ` `g.graph ``=` `[[``0``, ``1``, ``1``, ``1``], [``1``, ``0``, ``1``, ``0``], [``1``, ``1``, ``0``, ``1``], [``1``, ``0``, ``1``, ``0``]] ` `m ``=` `3` `g.graphColouring(m) ` ` `  `# This code is contributed by Divyanshu Mehta `

## C#

 `/* C# program for solution of M Coloring problem  ` `using backtracking */` `using` `System; ` ` `  `class` `GFG { ` `    ``readonly` `int` `V = 4; ` `    ``int``[] color; ` ` `  `    ``/* A utility function to check if the current  ` `    ``color assignment is safe for vertex v */` `    ``bool` `isSafe(``int` `v, ``int``[, ] graph, ` `                ``int``[] color, ``int` `c) ` `    ``{ ` `        ``for` `(``int` `i = 0; i < V; i++) ` `            ``if` `(graph[v, i] == 1 && c == color[i]) ` `                ``return` `false``; ` `        ``return` `true``; ` `    ``} ` ` `  `    ``/* A recursive utility function to solve m  ` `    ``coloring problem */` `    ``bool` `graphColoringUtil(``int``[, ] graph, ``int` `m, ` `                           ``int``[] color, ``int` `v) ` `    ``{ ` `        ``/* base case: If all vertices are assigned  ` `        ``a color then return true */` `        ``if` `(v == V) ` `            ``return` `true``; ` ` `  `        ``/* Consider this vertex v and try different  ` `        ``colors */` `        ``for` `(``int` `c = 1; c <= m; c++) { ` `            ``/* Check if assignment of color c to v  ` `            ``is fine*/` `            ``if` `(isSafe(v, graph, color, c)) { ` `                ``color[v] = c; ` ` `  `                ``/* recur to assign colors to rest  ` `                ``of the vertices */` `                ``if` `(graphColoringUtil(graph, m, ` `                                      ``color, v + 1)) ` `                    ``return` `true``; ` ` `  `                ``/* If assigning color c doesn't lead  ` `                ``to a solution then remove it */` `                ``color[v] = 0; ` `            ``} ` `        ``} ` ` `  `        ``/* If no color can be assigned to this vertex  ` `        ``then return false */` `        ``return` `false``; ` `    ``} ` ` `  `    ``/* This function solves the m Coloring problem using  ` `    ``Backtracking. It mainly uses graphColoringUtil()  ` `    ``to solve the problem. It returns false if the m  ` `    ``colors cannot be assigned, otherwise return true  ` `    ``and prints assignments of colors to all vertices.  ` `    ``Please note that there may be more than one  ` `    ``solutions, this function prints one of the  ` `    ``feasible solutions.*/` `    ``bool` `graphColoring(``int``[, ] graph, ``int` `m) ` `    ``{ ` `        ``// Initialize all color values as 0. This ` `        ``// initialization is needed correct functioning ` `        ``// of isSafe() ` `        ``color = ``new` `int``[V]; ` `        ``for` `(``int` `i = 0; i < V; i++) ` `            ``color[i] = 0; ` ` `  `        ``// Call graphColoringUtil() for vertex 0 ` `        ``if` `(!graphColoringUtil(graph, m, color, 0)) { ` `            ``Console.WriteLine(``"Solution does not exist"``); ` `            ``return` `false``; ` `        ``} ` ` `  `        ``// Print the solution ` `        ``printSolution(color); ` `        ``return` `true``; ` `    ``} ` ` `  `    ``/* A utility function to print solution */` `    ``void` `printSolution(``int``[] color) ` `    ``{ ` `        ``Console.WriteLine(``"Solution Exists: Following"` `                          ``+ ``" are the assigned colors"``); ` `        ``for` `(``int` `i = 0; i < V; i++) ` `            ``Console.Write(``" "` `+ color[i] + ``" "``); ` `        ``Console.WriteLine(); ` `    ``} ` ` `  `    ``// Driver Code ` `    ``public` `static` `void` `Main(String[] args) ` `    ``{ ` `        ``GFG Coloring = ``new` `GFG(); ` ` `  `        ``/* Create following graph and test whether it is  ` `        ``3 colorable  ` `        ``(3)---(2)  ` `        ``| / |  ` `        ``| / |  ` `        ``| / |  ` `        ``(0)---(1)  ` `        ``*/` `        ``int``[, ] graph = { { 0, 1, 1, 1 }, ` `                          ``{ 1, 0, 1, 0 }, ` `                          ``{ 1, 1, 0, 1 }, ` `                          ``{ 1, 0, 1, 0 } }; ` `        ``int` `m = 3; ``// Number of colors ` `        ``Coloring.graphColoring(graph, m); ` `    ``} ` `} ` ` `  `// This code is contributed by PrinciRaj1992 `

Output:

```Solution Exists: Following are the assigned colors
1  2  3  2
```

Complexity Analysis:

• Time Complexity: O(m^V).
There are total O(m^V) combination of colors. So time complexity is O(m^V). The upperbound time complexity remains the same but the average time taken will be less.
• Space Complexity: O(V).
To store the output array O(V) space is required.

References:
http://en.wikipedia.org/wiki/Graph_coloring

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